Question

The top of the pool table is 2.80 ft from the top of the floor. A pool player hits the ball and it lands 14.25 ft from the edge of the table. Calculate his break speed and the speed at which the ball hit the ground

Answer 1

This cannot be answered. There is just too much missing information to even guess at a possible answer.

There are multiple break speeds which can vary the vector of the cue ball after impact, each resulting in the cue ball landing 14.25 ft from the table. Here are two clear examples:

Example 1: The cue ball bounces off the rack just high enough to clear the rail and leave the table, landing 14.25 feet away.

Example 2: The cue ball bounces almost vertically into the air 25 feet (from a very hard break) and after returning from orbit, eventually lands a mere 14.25 feet from the table.

As you can imagine, each of these examples would have two completely different break speeds and landing speeds.

A long-winded explanation with more information:

The cue ball will bounce upward off the table only when the cue ball is slightly elevated (off the bed of the table) when it contacts the rack.

Why is the cue ball elevated during impact? The stick cannot hit the cue ball perfectly level because the rail of the table gets in the way. Therefore, the cue stick always imparts energy into the cue ball with a velocity vector that has a slight downward angle. This causes the cue ball to be forced into the table with a similar velocity as if it had been dropped or tossed onto the table. The result is that the cue ball will bounce off the surface of the table (usually very slightly) even though it started at rest on the table. This is common knowledge to advanced players because a legal jump shot in pool must be made by shooting the cue ball into the table (if you "scoop" under the cue ball to get it in the air, that's considered a foul.)

If that's true, why don't all breaks result in cue balls bounce upward from the break? A few reasons, but the main reason is because the cue ball might have landed from a bounce just as it reaches the rack. For advanced players, we know that if you're getting too much bounce then you're not putting all of the cue ball's energy into the rack, so we move the cue ball back a few inches before we break in an attempt to "adjust" where the bounce happens, hopefully just as it hits the rack.

When the cue ball is elevated off the bed of the table when it makes contact with the rack, then the velocity vector of the cue ball coming off of the rack will have some amount of elevation as well. The amount of elevation in this final vector is a product of the magnitude and angle of the incident vector at the moment of impact (there are other factors, such as the mass of the cue ball, friction, dirt on the balls, elasticity of the balls, etc. but we'll ignore them for simplicity.)

(I am a competitive pool player and the engineer that created a product called "Break Speed", which is a software program that is commonly used to track the break speeds of professional players during broadcast tournaments.)

Answer 2

[AI-generated answer — Explain Bot]

To calculate the break speed of the pool player, we can use the equation for horizontal motion:

𝑑 = 𝑣𝑥 × 𝑡

where 𝑑 is the horizontal distance traveled by the ball, 𝑣𝑥 is the horizontal velocity (break speed), and 𝑡 is the time of flight.

Given that the ball lands 14.25 ft from the edge of the table, we can assume the horizontal distance traveled is 14.25 ft. The time of flight, 𝑡, will be the same for both the horizontal and vertical motions.

Now, to calculate the time of flight, 𝑡, for the vertical motion (the time it takes for the ball to hit the ground), we can use the equation:

ℎ = 𝑣𝑦0 × 𝑡 − 0.5 × 𝑔 × 𝑡²

where ℎ is the vertical distance, 𝑣𝑦0 is the initial vertical velocity of the ball (which we can assume is zero since it starts from rest), and 𝑔 is the acceleration due to gravity (approximately 32.2 ft/s²).

Given that the top of the pool table is 2.80 ft from the top of the floor, the vertical distance traveled is 2.80 ft.

Now we can rearrange the equation for the vertical motion to solve for 𝑡:

2.80 = −0.5 × 32.2 × 𝑡²

Simplifying the equation gives:

14.25 = 𝑣𝑥 × 𝑡

Now we have two equations:

14.25 = 𝑣𝑥 × 𝑡
2.80 = −0.5 × 32.2 × 𝑡²

Solving the second equation for 𝑡, we find:

𝑡 = √(−2.80 ÷ (−0.5 × 32.2))

Using the obtained value of 𝑡, we can substitute it into the first equation to solve for 𝑣𝑥:

14.25 = 𝑣𝑥 × √(−2.80 ÷ (−0.5 × 32.2))

Simplifying this equation will give us the break speed, 𝑣𝑥, of the pool player.

To calculate the speed at which the ball hits the ground, we can substitute the value of 𝑡 into the equation for vertical motion:

𝑣𝑦 = 𝑣𝑦0 + 𝑔 × 𝑡

Considering 𝑣𝑦0 is zero, the equation simplifies to:

𝑣𝑦 = 𝑔 × 𝑡

Using the derived value of 𝑡, we can calculate the speed 𝑣𝑦 at which the ball hits the ground.

By implementing these steps, you can find both the break speed and the speed at which the ball hits the ground.